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i suppose first step would be to assume n is perfect square

so n = m^2

then i suppose i find n^2

i mean n + 2

n + 2 would be m^2 + 2

so now... i assume m^2 + 2 is a perfect square...

we have no 2 perfect square numbers with difference 2

so i let m^2 + 2 = k^2

@mukushla that's not really a proof.....

you may continue @igba

actually that is :)
\[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible

why so?

well,,you've almost proved it..
now m and k both are integers/.

and its k^2 -m^2 = 2 and not otherwise..

oh yes

2 = (k + m)(k-m)

now...integers huh...

\[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]

you know k>m..
so from
(k+m)(k-m)= 1*2
=>k+m =2
and
k-m = 1
this doesnt have an integral solution..

why do i know k > m again?

m^2 + 2 = k^2
both m and k are positive integers
hence k>m

in layman language,, k^2 - m^2 >0

when was it assumed that m and k are positive?

what else do you mean by perfect square ?

i don't see how that relates.... the definition of a perfect square is x = k^2

4 is a perfect sq since its 2^2 i.e. square of a positive integer..
and likewise..

(-2)^2 is also 4...

you take the absolute value

i mean you have to..

why?

...i don't think this is a valid proof....

since when did sqrt x become |-y|

0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats.
No +2 in there.

so and i think that this difference of two will be the ,,key" of this proof of ...

@lgbasallote what is your opinion from this ,please ?

my opinion is that this is some tricky algebra

how do you think it ???

so and why ?

because i can't think of a proof

how can you prove/disprove that the difference of two squares is 2...

hmmm...wait...

their fields of specialty aren't exactly in proving

What's wrong with my proof?

and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..

@estudier i mean

@mukushla too

So, that doesn't mean it's invalid..

it does actually.....you can't do infinite substitutions to prove anything....

There is no infinity involved, all integer squares end in the numbers I gave.

infinite loops involves infinity agree?

No infinite loop either, just logic.

An integer ends in one of digits 0 to 9.
The squares must end in etc....

but we both know that setting values is illegal in proving

I am not setting any values.

and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever

you're checking the squares of each number

I am not.

and though there is a pattern...there is still susbtitution involved

and then you'll have to substitute forever to verify that the pattern really does not break

then infinity is involved

It's not a pattern, it's a fact
0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1

so you do admit that you checked the 0 to 9 squares

Anything else is impossible.

...that's substitution

anyway...my point is...substitution isn't really allowed in proving

You will find that this fact is used a lot in many valid proofs in number theory.

perhaps

OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.

can you prove it? that it's a fact about squares?

Yes, all numbers end in 0 to 9 and so their squares end in .....QED.

you didn't really demonstrate anything...you just repeated what you said

You asked for proof, I just gave u a proof.

If you deny my proof, provide a counterexample.

since you try so hard to defend your proof, i'll accept it

That's very gracious, thank you:-)

OK, I will look at it now....

again i ask... how is k > m

(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

again i ask for the nth time @mukushla how is k > m

ok man i just want to answer ur question

well because 8 - 4 is not 2 but it's even minus even

@mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?

OK, so finally, the algebra
Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

i thought they are positive :) sorry

why (k+1)^2 @estudier ?

I already, said, that's the next square....

next square?

is this still related to my proof?

? n is k^2, what is the next square?

No, this proof
Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

are you allowed to do that? @mukushla

can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end

why not...for any real number \[a^2=|a|^2\]

Given n=k^2 ->
(k+1)^2 = k^2 + 2k +1
k^2 + 2k +1 >= n +3
n +3 > n+2 QED

That's 2 proofs I have given now...

Think about it....:-)

in the meanwhile... @mukushla why |k| - |m| = 1?

because\[|k|-|m|<|k|+|m|\]

how does that make |k| - |m| = 1

right?

who said those are positive?

|k|>|m| right?

something feels wrong...

anyway...where did you get 2|k| > 3

ok...think on what i said...because i believe thats the shortest way to prove ur statement

|2k| = 3 *

adding\[|k|-|m|=1\]and\[|k|+|m|=2\]

Going back to your question about 2, u can just adjust it to difference of 2q...

im really skeptical about this... especially that |k|^2 - |m|^2 = 2

difference of 2q?

You were worried about the validity of your statement, change it to 2q instead of just 2.

Or 2p, or whatever letter u like best.

oh. when i said difference of two numbers equal to 2

but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?

and @mukushla how did you get |k| > |m|

Yes, the algebra is completely general...

i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...

...which is greater than n+3, right?

it is?

Number theorists don't pay attention to 0

unless k is 0....

why not?

Can I try?

More the merrier.

"because i know nothing about it"
But you do now:-)

it's not like it's limited to the 99s....

Yes, that's similar to the one I gave...

"number theorists don't consider 0"
http://en.wikipedia.org/wiki/Natural_number

http://mathforum.org/library/drmath/view/63510.html
Using 0,1,4, etc...